# Nature of a wave

We have two transverse wave

$$y_1= A\sin(kx-\omega t)$$

$$y_2=-A\sin(kx+\omega t)$$

These two waves superposed. Then, at point $x=0$ the equation would be

$$y_1 + y_2 =-2A\sin\omega t.$$

Now how can we know at $x=0$ is node, antinode or neither of them .

In my book it is given as antinode .

• @JohnRennie yes they are moving in opposite direction – user123733 Jan 4 '17 at 17:14
• It appears you have made a mistake in your solution. Please see hint below. – InertialObserver Jan 4 '17 at 17:24

Hint(1): \begin{align} \sin(kx-\omega t)+\sin(kx+\omega t)&=\sin(kx)\cos(\omega t)-\cos(kx)\sin(\omega t)+\sin(kx)\cos(\omega t)+\cos(kx)\sin(\omega t)\\ &=2\sin(kx)\cos(\omega t) \end{align}

Hint(2):

By definition, the nodes are when the quantity $\sin(kx)=0$.

Spoiler! Hover over yellow for answer.

That is, they are located at the $x_n$ such that $$kx_n=n\pi$$ where $n=0,\pm 1,\pm 2\cdots$. That is, $$x_n=\frac{n\pi}{k}$$ are where the nodes are located.

• But why we can add like , when x=0 $y_1=Asin(- \omega t)$ and $y_2=-Asin\omega t$ and add hoth of them . – user123733 Jan 4 '17 at 17:28
• In my book it is given as antinode. How ? – user123733 Jan 4 '17 at 18:21
• Sorry , I have edited my question – user123733 Jan 4 '17 at 18:24
• @user123733 You completely changed the question, so you changed the solution from node to antinode. Given your equation -2Asin(wt) at x=0, it should be obvious that it's an antinode since the y position changes sinusoidally with an amplitude of the sum of the two other waves. This means x=0 is a point where maximum displacement occurs, which is the condition for an antinode. – Steve Jan 5 '17 at 3:18
• @user123733 Yes the position of an antinode changes with time, it doesn't stay fixed at the maximum, it only takes on that maximum value at specific times. A node stays fixed in place, but an antinode moves the largest possible amount over time. There are likely some YouTube videos that can demonstrate an antinode visually over time. – Steve Jan 5 '17 at 3:56